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The directions for this problem are to determine whether or not the sequence converges or diverges, and to calculate the limit if it exists.

I know that the only thing you can really do with sequences for this is to take the limit, but I am not sure how to go about calculating this limit. I tried manipulating it so it would be infinity over infinity for L'Hopital's rule, but that didn't work. The sequence is

$$A_n = (-1)^nn^2sin(\frac{1}n)$$

I am right now trying to solve

$$\lim\limits_{n \to \infty} (-1)^nn^2sin(\frac{1}n)$$

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    $\begingroup$ Do you know that $n^2sin(1/n)\rightarrow \infty$? can you use that? $\endgroup$ – Yanko Nov 15 '17 at 19:29
  • $\begingroup$ Do you know that as $x\to 0$, $sin(x) \approx x$? $\endgroup$ – Paolo Intuito Nov 15 '17 at 19:31
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You can use this

$$\left\vert(-1)^nn^2\sin(\frac{1}n)\right\vert=n\frac{\sin(1/n)}{1/n}\sim n\xrightarrow{}\infty$$

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